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3 years ago

12

Which equation describes the same line as y- 6 = -4(x + 1)?

2 answers:

just olya [345]3 years ago

8 0

Answer:

B. y=-4x+2

Step-by-step explanation:

y-6=-4x-4

y=-4x-4+6

y=-4x+2

NemiM [27]3 years ago

3 0

The answer is B. y= -4x+ 2

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Dan is cutting 4.75 foot length of twine from a 240 foot spoil of twine. He needs to cut 42 lengths and says that 40.5 feet of t

Bad White [126]

1 length is 4.75 feet

42 lengths = x feet Cross multiply

1 * x = 42 * 4.75 Multiply on the right

x = 199.5

amount left = amount started with - amount amount used

Amount left = ???

Amount started with = 240 feet.

Amount used = 199.5 Substitute

Amount left = 240 - 199.5 Subtract

Amount left = 40.5 Answer

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3 years ago

Using point-slope form, write the equation of the line that passes through (2, –1) and has a slope of –5.

erik [133]

Answer: (y+1)=5(x+2)

Step-by-step explanation:

6 0

3 years ago

URGENT!!!! PLEASE HELP NOW!!! WHO EVER GIVES THE CORRRECT ANSWER WILL GET BRAINLIEST!!!

sladkih [1.3K]

Answer:

The correct ans is..... ( which i believe )

3rd option

Hope this helps...

Pls mark my ans as brainliest

If u mark my ans as brainliest u will get 3 extra points

3 0

3 years ago

Read 2 more answers

Shtirlitz [24]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Here's the solution ~

As we know, we can calculate the circumference of a circle in terms of its diameter as :

$\qquad \sf \dashrightarrow \:c = \pi d$

where, c = circumference and d = diameter

And also, circumference of circle is terms of radius (r) is :

$\qquad \sf \dashrightarrow \:c = 2\pi r$

Now, let's move on to questions ~

<h3>First </h3>

$\qquad \sf \dashrightarrow \:3.14 \times 5.9$

$\qquad \sf \dashrightarrow \: \approx18.53 \: ft$

<h3 />

・．━━━━━━━†━━━━━━━━━．・

<h3>Second</h3><h3 /><h3 /><h3 /><h3> $\qquad \sf \dashrightarrow \:3.14 \times 3.2$ </h3>

$\qquad \sf \dashrightarrow \: \approx10.048 \: ft$

・．━━━━━━━†━━━━━━━━━．・

<h3>Third</h3>

$\qquad \sf \dashrightarrow \:3.14 \times 6.1$

$\qquad \sf \dashrightarrow \: \approx19.15 \: ft$

・．━━━━━━━†━━━━━━━━━．・

<h3>Fourth</h3>

$\qquad \sf \dashrightarrow \:2×3.14 \times 3.7$

$\qquad \sf \dashrightarrow \: \approx2×11.62 \: m$

$\qquad \sf \dashrightarrow \: \approx23.24 \: m$

・．━━━━━━━†━━━━━━━━━．・

<h3>Fifth </h3>

$\qquad \sf \dashrightarrow \:2×3.14 \times 6.2$

$\qquad \sf \dashrightarrow \: \approx2× 19.47 \: units$

$\qquad \sf \dashrightarrow \: \approx38.94 \: m$

・．━━━━━━━†━━━━━━━━━．・

<h3>Sixth</h3>

$\qquad \sf \dashrightarrow \:2×3.14 \times 5.1$

$\qquad \sf \dashrightarrow \: \approx2×16.01 \: ft$

$\qquad \sf \dashrightarrow \: \approx \: 32.02m$

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

7 0

2 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago